A digression on Hermite polynomials

# A digression on Hermite polynomials

KEITH Y. PATARROYO
kypatarroyot@unal.edu.co
July 20, 2019

Orthogonal polynomials are of fundamental importance in many fields of mathematics and science, therefore the study of a particular family is always relevant. In this manuscript, we present a survey of some general results of the Hermite polynomials and show a few of their applications in the connection problem of polynomials, probability theory and the combinatorics of a simple graph. Most of the content presented here is well known, except for a few sections where we add our own work to the subject, nevertheless, the text is meant to be a self-contained personal exposition.

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## 1 A general overview of the polynomials

We start with the definition of the polynomials and some details regarding the notation. Afterward, we pursue a construction and an explicit expression for them.

### 1.1 Definition

The Chebyshev111We are calling the functions , Chebyshev-Hermite polynomials following the convention of [2] and [3] to differentiate them from (see Note 1 in the text). However they are most commonly known by Hermite polynomials alone(hence the title of this article), e.g. see [4],[5],[6]. It is our hope that "Chebyshev-Hermite polynomials" becomes generally accepted since at the present time it is only known by some people in the statistical community [7]. Also, note that these polynomials are NOT the Hermite-Chebyshev polynomials defined in [8].-Hermite polynomials arise naturally when we take the ratio of two standard Gaussian distributions[6], this motivates the following definition.

###### Definition 1

The Chebyshev-Hermite polynomials are defined in the interval and they are generated by the ratio of a displaced standard Gaussian distribution and a standard Gaussian distribution centered in zero,

 ϕ(x−t)ϕ(x)=ext−t22=∞∑n=0Hen(x)tnn!=G(x,t). (1)

This definition tells us how to compute the polynomials and it shows that the Chebyshev-Hermite polynomials have to some extent a connection with the Gaussian distribution. Furthermore we can see right away that this generating function (1) is quite similar to the moment generating function of a random variable with normal distribution. A substantial relation of with and will arise when we consider the integral representation and the connection problem of the polynomials in Sections 2.5 and 3.1 respectively.

###### Note 1

Hermite polynomials are standardized in two different ways depending on their use, they are: the Chebyshev-Hermite polynomials, (regularly applied in probability [6]) and the Hermite polynomials, (frequently applied in physics [9],[10]). Sometimes the naming of the polynomials change from author to author, however the mathematical notation for both is currently well distinct in the technical literature [4]. They both share the following relationship,

 Hn(x)=2n2Hen(√2x). (2)

### 1.2 Construction by a Gram-Schmidt algorithm

One way to naturally obtain classical222 The Chebyshev-Hermite polynomials are a classical orthogonal polynomial sequence, Why classical? Are there any other? e.g. quantum?, see [10] or [11] for a precise definition. orthogonal polynomials is by constructing a set of polynomials with such that they are orthogonal[4] under an inner product with a weight function on the interval . The following theorem guarantees that such set of polynomials exist and are unique, moreover it tell us a way to iteratively compute them.

###### Theorem 2

Given an inner product with a weight function on an interval , there exists a sequence of polynomials orthogonal with respect to the inner product. This sequence is uniquely determined up to constant factors333As we will see later the Chebyshev-Hermite polynomials are monic[4], Exercise 1, and according to Theorem 2 they are only undetermined up to a constant factor. Now lets consider , these polynomials are monic too !! What is wrong? Is the weight function for equal to the weight function for ?.

Proof: See [12] for complete details on the proof. Here we give an outline on the proof, this requires constructing the polynomials using a Gram-Schmidt like algorithm to orthogonalize the linear independent sequence using a determined inner product with a weight function . This construction is not the most efficient approach to build the polynomials, since it requires lots of computations, see [12] for an example of this procedure, later we will see better constructions algorithms.

If we choose the weight function to be an unnormalized standard Gaussian distribution, 444Why choose as a weight function anyway? For an arbitrary function to be a weight function, it must be positive, continous and all its moments should exist [15]. In the other hand the classification of classic orthogonal polynomials [10] yields that a function of the form must generate a sequence of classic orthogonal polynomials.  and the interval , we obtain the sequence of Chebyshev-Hermite polynomials as the set of orthogonal polynomials, the orthogonality is proven later in Lemma 4. Hence if we are working in an application where the inner product has the alleged weight function, the natural way to write an arbitrary function, as a series of orthogonal functions, is using the Chebyshev-Hermite polynomials. We can do this since the Chebyshev-Hermite polynomials are a Complete Orthogonal System in [13].

### 1.3 Explicit expression

A remarkable, however vague result is a general explicit expression for the Chebyshev-Hermite polynomials. To obtain it we start with the generating function (1) and expand the exponential argument,

 ext−t22=∞∑k=0(xt−t2/2)kk!=∞∑k=0k∑j=0(kj)(xt)k−j(−1)j(t)2j2jk!,

where we used the binomial theorem to obtain the last expression. Now we redefine indexes, so we let and rewrite both sums,

where is the floor function. Now comparing with (1) we find that555 An explicit expression is wonderful for calculating the polynomials in a computer, however, it is cumbersome (yet (4) and (3) will be really important in some applications, Section 3), it is hard to see some properties of the polynomials with the explicit expression. As we will see later other generating methods, Section 2 (recurrence relation, Rodrigues formula, ODE, integral representation) may highlight the importance of a property more than another. This is maybe why orthogonal polynomials are so powerful, one is able to apply different perspectives to solve a problem.,

 Hen(x)=n!⌊n2⌋∑j=0(−1)jxn−2j2j(n−2j)!j!. (3)

We will later need in this document the explicit expression for the Hermite polynomials, , so using equation (2) we can write their explicit expression,

 Hn(x)=n!⌊n2⌋∑j=0(−1)j(2x)n−2j(n−2j)!j!. (4)
###### Exercise 1

Check the following special values for the Chebyshev-Hermite polynomials:

• Leading Coefficient: , then the polynomials are monic[4].

• Final Coefficient: and .

• Parity : .

The reader is encouraged to prove some of these properties with the other generating methods for the polynomials, Sections 1.1, 1.2 and Section 2. This will make him more comfortable with the different representations of the polynomials.

With the explicit expression (3) we can make a table for the first six Chebyshev-Hermite polynomials and a plot of them. We present this data in the following figure,

We can rapidly check the properties of Exercise 1 with the table in Figure 1. Also, from the figure we see other beautiful properties of the polynomials, the polynomial of order has real roots and between any two roots of a polynomial there is a root of a higher order polynomial, these facts will be of use later.

## 2 Some fundamental results

We proceed next to present some features of the Chebyshev-Hermite polynomials that are commonly shown in the treatment of orthogonal polynomials.

### 2.1 Recurrence relation

All classical orthogonal polynomials follow a recurrence relation as is well known in the technical literature [4], in some treatments of orthogonal polynomials the generating function is derived from the recurrence relations [10]. We do the opposite process and derive two recurrence relations from the generating function (1).

###### Lemma 3

The sequence of the Chebyshev-Hermite polynomials, , satisfy the two following recurrence relations,

 He′n(x)=nHen−1(x), for n≥1, (5)
 Hen+1(x)−xHen(x)+nHen−1(x)=0, for n≥1. (6)

Proof: We start with the generating function (1), and we note that it satisfies the following two differential equations,

 ∂G∂x(x,t) = text−t22=tG(x,t), ∂G∂t(x,t) = xext−t22−text−t22=(x−t)G(x,t).

Now if we plug the series expansion for the generating function (1) in the previous differential equations, we can obtain equalities relating the coefficients of the series. First for the derivative with respect to we have,

 ∂G∂x(x,t)=∞∑n=1He′n(x)tnn!=∞∑m=1mHem−1(x)tmm!=∞∑n=0Hen(x)tn+1n!=tG(x,t),

where we made in the third equality, now equating for the coefficients of the series we find (5). Finally the derivative with respect to yields,

 ∂G∂t(x,t)=∞∑n=1nHen(x)tn−1n!=∞∑k=0Hek+1(x)tkk! = ∞∑n=0xHen(x)tnn!−∞∑m=0mHem−1(x)tmm!, = ∞∑n=0xHen(x)tnn!−∂G∂x(x,t), = (x−t)G(x,t),

where we took the series expansion of from the previous manipulation. Now equating for the coefficients of the series we find (6). This proves the lemma.

Now we present an example showing the power of this recurrence relation. We will show later another solution for this example in Section 3.1.

###### Example 1

We compute the convolution of a standard Gaussian distribution with the nth Chebyshev-Hermite polynomial , also called the rescaled Weierstrass-Gauss transform [16],[17] of the Chebyshev-Hermite polynomial ,

 W[2n2Hen(x√2)]=(ϕ∗Hen)(x)=∫∞−∞(2π)−1/2e−(y−x)22Hen(y)dy.

We claim that , we see that the convolution yields the moments of a random variable with normal distribution. Then we clearly notice from the table in Figure 1 that for ,

 (ϕ∗He0)(x)=E[Y0]=1,(ϕ∗He1)(x)=E[Y1]=x.

Now we employ induction on , suppose then for we have,

 (ϕ∗Hen−1)(x)=∫∞−∞(2π)−1/2e−(y−x)22Hen−1(y)dy=xn−1. (7)

Consider now and lets use both (5),(6) with ,

 ∫∞−∞(2π)−1/2e−(y−x)22Hen(y)dy=∫∞−∞(2π)−1/2e−(y−x)22(yHen−1(y)−He′n−1(y))dy. (8)

We can then compute the integral with the term by integration by parts,

 ∫∞−∞(2π)−1/2e−(y−x)22He′n−1(y)dy=∫∞−∞(2π)−1/2(y−x)e−(y−x)22Hen−1(y)dy,

where we took into account that multiplied by any polynomial vanish for infinite . Now replacing the last expression in (8) we obtain,

 ∫∞−∞(2π)−1/2e−(y−x)22Hen(y)dy=x∫∞−∞(2π)−1/2e−(y−x)22(Hen−1(y))dy.

Finally using the induction hypothesis (7) in the last expression we obtain,

 W[2n2Hen(x√2)]=E[Hen(Y)]=(ϕ∗Hen)(x)=xn, (9)

which is the desired result.

###### Exercise 2

Compute the integral (9) starting from the generating function (1), multiply both sides by integrate in and compare the terms in the series.

### 2.2 Rodrigues formula

Now we derive the so-called Rodrigues formula for the Chebyshev-Hermite polynomials, this formula is extremely useful to solve many problems quickly. We start with our generating function (1), we write it in a convenient way and recognize the Taylor expansion coefficients as,

 [∂n∂tn(ext−t22)]t=0=ex22[∂n∂tn(e−(t−x)22)]t=0=Hen(x).

Using the identity in the above expression yield,

 ex22[∂n∂tn(e−(t−x)22)]t=0=ex22(−1)n[∂n∂xn(e−(t−x)22)]t=0=(−1)nex22dndxn(e−x22).

Then the Rodrigues formula for the Chebyshev-Hermite polynomial is,

 Hen(x)=(−1)nex22dndxn(e−x22). (10)
###### Exercise 3

Show that the Chebyshev-Hermite polynomials can also be defined in the following two ways,

 Hen(x)=ex24(x2−ddx)n⋅[e−x24],Hen(x)=(x−ddx)n⋅1. (11)

Now we prove the orthogonality of Chebyshev-Hermite polynomials, we decide to present this proof since it shows the power of the Rodrigues formula. Again the reader is encouraged to prove this with the other generating methods for the polynomials.

###### Lemma 4

The Chebyshev-Hermite polynomials are orthogonal with respect to the weight function in the interval . In other words,

 ∫∞−∞e−x22Hen(x)Hen′(x)dx=√2πn!δnn′. (12)

Proof: We start by expressing with the Rodrigues formula, (10), then the integral yields,

For the case assume without loss of generality that , now we integrate by parts times the above expression, we take into account the fact that and all its derivatives vanish for infinite ,

 ∫∞−∞e−x22Hen(x)Hen′(x)dx = (−1)n−1∫∞−∞dn−1dxn−1(e−x22)He(1)n′(x)dx, = ⋮ = (−1)n−n′∫∞−∞dn−n′dxn−n′(e−x22)He(n′)n′(x)dx, = 0,

where its clear that the derivative of a degree polynomial is zero.

For the case , we perform the same procedure as above, we integrate by parts times ,

 ∫∞−∞e−x22Hen(x)Hen(x)dx = (−1)0∫∞−∞d0dx0(e−x22)He(n)n(x)dx, = n!∫∞−∞e−x22dx, = n!√2π,

where we made use of the leading coefficient of the Chebyshev-Hermite polynomials, Exercise 1, and the Gaussian integral. This proves the lemma.

Now a little exercise where one can apply the Rodrigues formula and the method applied above to prove Lemma 4 of successive integration by parts.

###### Exercise 4

Obtain the inverse explicit expression from (3) for the Chebyshev-Hermite polynomials,

 xn=n!⌊n2⌋∑j=0Hen−2j(x)2j(n−2j)!j!. (13)

Suggestion: Divide in two cases and , then expand the power as a linear combination of Chebyshev-Hermite polynomials. Use orthogonality to find the coefficients, do the necessary integrals with the process described above. Join the even and odd result to find (13).

We will later need in this document the inverse explicit expression for the Hermite polynomials, , so using equation (2) and (13) we can write their inverse explicit expression from (4) as,

 (2x)n=n!⌊n2⌋∑j=0Hn−2j(x)(n−2j)!j!. (14)

Next, we apply some of the results we obtained in the previous sections to a very important application of the polynomials, numerical integration.

### 2.3 Gauss-Hermite Quadrature

Imagine one would like to compute a very difficult integral in the interval , since the interval is unbounded, the direct application of classical techniques (trapezoid or unmodified Monte Carlo) for the numerical computation of the integral are unsuccessful. Remarkably the Chebyshev-Hermite polynomials can be used to tackle this problem directly using the so-called numerical quadrature or Gauss quadrature.

The quadrature rule for a general family of orthogonal polynomials with respect to a weight on an interval is the result of following theorem,

###### Theorem 5

Let be a degree polynomial and be the zeros of , then the following quadrature formula holds,

 ∫baw(x)f(x)dx=N∑i=1wi,Nf(xi),wi,N=aNaN−1⋅∫baw(x)[pN−1(x)]2dxp′N(xi)pN−1(xi), (15)

where is the coefficient of the orthogonal polynomial .

Proof: See [14] for details.

It can also be proven, see [14], that all the zeros of for are real, distinct and lie in , hence (15) is always computable.

If is not a polynomial, (15) is not exact, roughly the discrepancy is a result of how well is approximated by a polynomial. However if is continuous the approximation gets better as we increase the number of quadrature points [15].

For the particular case of the Chebyshev-Hermite polynomials, using equations (12), (5) and Exercise 1 we obtain,

 ∫∞−∞e−x22f(x)dx=N∑i=1wi,Nf(xi),wi,N=√2πN![NHeN−1(xi)]2, (16)

where are the zeros of .

We can use the Chebyshev-Hermite polynomials to solve the original problem of a difficult integral of the form for arbitrary complicated , one way to do this is write as (some approaches are described in Section 3.2.1), then the integral can be approximated as,

 I=∫∞−∞e−x22^f(x)dx≈N∑i=1wi,N^f(xi).

Even though this method might require two approximations, one can quickly improve the estimate by a high order quadrature rule since the procedure is just a matter of evaluate in some points, multiply by (both the points and weights can be stored in memory) and sum the outcomes.

### 2.4 Differential Equation(ODE)

The Chebyshev-Hermite polynomials also arise in many fields since they satisfy the eigenvalue problem,

 u′′−xu′=−nu, (17)

which for positive integer is widely regarded as the Hermite equation.

Its possible to solve this equation by the power series method and show that one family of solutions is indeed the Chebyshev-Hermite polynomials, [18]. However here we only show that the Chebyshev-Hermite polynomials satisfy (17), this is summarized in the following lemma,

###### Lemma 6

The sequence of the Chebyshev-Hermite polynomials, , satisfy the Hermite differential equation (17).

Proof: We start by replacing the recurrence relation (5) in the recurrence relation (6), this yields,

 Hen+1(x)−xHen(x)+He′n(x)=0.

Now taking the derivative of the previous equation we obtain,

 He′n+1(x)−Hen(x)−xHe′n(x)+He′′n(x)=0.

Using (5) with the term in the previous equation, we obtain,

 He′′n(x)−xHe′n(x)=−nHen(x). (18)

This proves the lemma.

Its convenient to introduce the orthogonal Chebyshev-Hermite functions , defined by,

 hen(x)=e−x24Hen(x). (19)

In the following exercise the reader is encouraged to check some elementary properties of the Chebyshev-Hermite functions,

###### Exercise 5

Show that the Chebyshev-Hermite functions are orthogonal in (with weight function 1) and they satisfy,

 x2hen(x)+he′n(x)=nhen−1(x), (20)
 he′′n(x)+(−x24+n+12)hen(x)=0. (21)

Equation (21) is known as Weber equation [19], it comes up in the study of the Laplace’s equation in parabolic coordinates; for arbitrary , the solutions of this equation are known as parabolic cylinder functions.

Another remarkable set of functions are the Hermite functions , they share a similar relation to (2) with the Chebyshev-Hermite functions given by,

 hn(x)=2n2hen(√2x)=e−x22Hn(x). (22)

Just as the Chebyshev-Hermite functions, the Hermite functions are orthogonal in and they satisfy the following differential equation(also called Hermite equation),

 h′′n(x)−x2hn(x)=−(2n+1)hn(x), (23)

as it can be directly checked or derived from (21).

The Hermite functions play a fundamental role in Quantum Mechanics (see [9],[10]) and in Fourier analysis as we will see on Section 2.6.

### 2.5 Integral representation

We present yet another way of obtaining the Chebyshev-Hermite polynomials. This time, the method depends on the remarkable result that is its own Fourier transform.

In other words, computing the inverse Fourier transform of is , this is (using the Fourier Transform defined in [10]),

 e−x22=F−1[e−k22]=1√2π∫∞−∞e−k22eixkdk. (24)

Since this result is fundamental for the rest of the section, we present a derivation in the following example.

###### Example 2

We compute the following integral,

 I(x)=1√2π∫∞−∞e−k22eixkdk.

Clearly the previous integral is equal to the following integral,

 I(x)=2√2π∫∞0e−k22cos(xk)dk. (25)

Taking the derivative of with respect to and integrating by parts we get,

 dI(x)dx = −2k√2π∫∞0e−k22sin(xk)dk, = −x[2√2π∫∞0e−k22cos(xk)dk].

Then the integral, satisfies the ordinary differential equation,

 dI(x)dx=−xI(x).

The solution of this ODE is , to find , we evaluate (25) in ,

 C=I(0)=2√2π∫∞0e−k22dk=1.

So we found that,

 I(x)=e−x22.
###### Exercise 6

Compute the integral (25) using (a) contour integration, and (b) by expanding in powers of and integrating term by term.

Now if we differentiate the integral (25) times respect to we get,

 dndxn(e−x22)=(i)n√2π∫∞−∞kne−k22eixkdk.

Now inserting the previous result in the Rodrigues formula (10) we find the integral representation of the Chebyshev-Hermite polynomials,

 Hen(x)=ex22(−i)n√2π∫∞−∞kne−k22eixkdk. (26)

From the previous result we can easily find the following Fourier transforms,

 F−1[kne−k22]=inHen(x)e−x22,F[xne−x22]=(−i)nHen(k)e−k22. (27)

With all the previous tools in hand, now we are ready to understand the relation of the raw moments from a random variable with normal distribution and the Chebyshev-Hermite polynomials.

###### Example 3

In spirit of [20] we present a remarkable way to compute the moments of a random variable with a general Gaussian distribution. For this we rewrite the integral representation (26) as follows,

 Hen(x)(−i)n=(−1)n√2π∫∞−∞kne−(k−ix)22dk=1√2π∫∞−∞zne−(z+ix)22dz,

where we used the change of variable in the last step. Now we let in the last equation and we recognize the raw moments from a random variable with normal distribution,

 Hen(i^x)(−i)n=1√2π∫∞−∞zne−(z−^x)22dz=E[Yn]. (28)

So we solved part of the mystery of the coefficients of the raw moments of , they are they are all positive as can be seen from (3), in fact they are the absolute value of the coefficients of the Chebyshev-Hermite polynomials. A simpler way to see this is by doing the change and in (1) obtaining precisely the moment generating function of a random variable .

Now scaling properly (28), we can find the the raw moments, using and their explicit expression, the details are left as an exercise for the reader,

 E[^Yn]=(−iσ)nHen(iμ/σ),E[^Yn]=σnn!⌊n2⌋∑j=0(μ/σ)n−2j2j(n−2j)!j!. (29)

After a long road of working with Chebyshev-Hermite polynomials and Chebyshev-Hermite functions, we could not resist adding a section entirely to the remarkable Fourier transform of Hermite functions (22). This deserves to be presented to encourage the reader to learn more about the incredible field of Fourier analysis.

### 2.6 Fourier Transform of Hermite functions*

* This section can be omitted without loss of continuity. It is meant for the ambitious reader or as an interesting general reference.

The Fourier transform is an essential tool in many areas of math and science, we present here the outstanding result that the Hermite functions (22), are the eigenfunctions of the Fourier Transform integral operator.

To informally see this, we take the Hermite equation and its Fourier transform, lets write equation (23) using an arbitrary function ,

 f′′(x)−x2f(x)=−(2n+1)f(x).

Using elementary results from Fourier transform theory [21], we see that the Fourier transform of this equation is,

 ^f′′(k)−k2^f(k)=−(2n+1)^f(k),

where we denoted , and took from the table in [21].

This means that both the function and its Fourier Transform satisfy the same differential equation, (23). One possible option can be that and its Fourier Transform are both Hermite functions and proportional666In fact more technically, this means that preserves each eigenspace of the differential operator , moreover the Fourier transform operator conspires with the space to restric the possibilites and make the only solution that and its Fourier Transform are both Hermite functions and proportional, see [22] for more detalis.. However we will see in a moment that this is not the complete answer.

Now we proceed to find the Fourier transform of the Hermite functions and to check they are effectively the eigenfunctions of the operator. For a more complete and remarkable treatment of the eigenproblem of the Fourier Transform see [22]. Coming back to our approach first note that by the Fourier inversion theorem, applying twice the Fourier operator yields , then .

Next to gain some insight, we consider momentarily the eigenvalue problem for the Fourier linear integral operator, . Then if we apply three more times to we get,

 f(k)=I[f](k)=F4[f](k)=λ4f(k).

This forces , so the eigenvalues are . In fact even functions have -eigenvalues and odd functions have -eigenvalues .

Now we use the generating function of the Hermite functions to find their Fourier transform. First note that the function is the generating function for the Hermite polynomials (2), this can be easily seen by reproducing the procedure in Section 1.3 with and show that one arrives to (4). Hence the generating function for the Hermite functions is,

 e(−x22+2xt−t2)=∞∑n=0hn(x)tnn!. (30)

Then we can rewrite the right hand side of the previous equation,

 e(−x22+2xt−t2)=e(−x2+4xt−4t22+2t22)=e−(x−2t)22⋅et2.

Applying the Fourier transform to (30), the right hand side yields,

 F[e(−x22+2xt−t2)]=et2F[e−(x−2t)22]=et2e−k22e−2ikt=e(−k22+2k(−it)−(−it)2),

where we used the result for a Fourier transform of a Gaussian function (24) and the space shifting property for the Fourier transform [21].

We recognize this Fourier transform as the generating function (30) with and , then the Fourier transform of the right and left hand side yields,

 ∞∑n=0hn(k)(−i)ntnn!=F[e(−x22+2xt−t2)]=F[∞∑n=0hn(x)tnn!]=∞∑n=0F[hn(x)]tnn!.

Now equating for the coefficients of the series we find,

 F[hn](k)=(−i)nhn(k). (31)

Then we see more clearly that the Hermite functions solve the eigenvalue problem with eigenvalues .

To comprehend why this is the only solution of the eigenvalue problem, see the derivation from [22], in addition, another useful source is [23]. It turns out this remarkable result has some applications, like proving the Heisenberg’s inequality (uncertainty principle)[15], also one can prove Plancherel formula with Hermite polynomials, see [24] and references therein.

### 2.7 Hermite polynomials in higher dimensions

One can generalize the single variable Chebyshev-Hermite polynomials in various ways. An option is to construct a two-parameter family of single variable orthogonal polynomials as is done [25], this is useful for generalizations of the Fourier transform or the quantum harmonic oscillator algebra [26]. Also, other interesting generalizations are [8],[27],[28]. Now we focus on the generalization of the polynomials to higher dimensions or multiple variables.

If we take the Rodrigues formula (10), we recognize that it can be written as follows777One can also start from (32) with different weight functions to construct different families of classical orthogonal polynomials, this is the approach taken by [10].

 Hen(x)=(−1)nw(x)dndxn[w(x)], (32)

where is the weight function of the Chebyshev-Hermite polynomials.

Now using the usual standard tensor notation, very much like the manuscript [29], let , we can extend both the definition of the weight function and the Chebyshev-Hermite polynomials as follows,

 w(x)=e−∥x∥22,He(n)(x)=(−1)nw(x)∇(n)[w(x)], (33)

where both and are tensors of rank , which can also be written in index notation as where are a list of indexes symbolizing any of the space coordinates.

###### Exercise 7

Check the following special cases for the first -dimensional Chebyshev-Hermite polynomials,

 He(0)(x) = 1, He(1)α1(x) = xα1, He(2)α